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\begin{document}
\section{Leftovers}

Is this true??
\begin{lem}\label{lem:cond}
Consider random variables $X,Y$ such that $X \le_{st} Y$. Then for all $a$, we have that 
\begin{align} \label{eq:conditional}
[X | X \ge a] \le_{st} [Y | Y \ge a].
\end{align}
\end{lem}
No, this is wrong \Frowny{}. According to \cite{shaked2007stochastic}, page 17, the equation \ref{eq:conditional} is equivalent to $X \le_{hr} Y$ which is not implied by $X \le_{st} Y$.

 Before that, we prove two auxiliary lemmas.
\begin{lem}
Let $X_a$, and $Y_a$ be random variables ordered by $a$, with hazard rate functions $r_a$, and $q_a$, and mean $l_a$, and $m_a$, respectively. Provided $r(t) \ge q(t)$, and $q(t)-r(t) \downarrow t$, we have that $l(t) \le m(t)$, and $m(t)-l(t) \downarrow t$.
\end{lem}
\begin{lem}
Let $X$ be an IFR random variable, and $f$ be an arbitrary increasing function. Denote the residual lifetime of $X$ at $t$ by $X(t)$. Then, $\Ex f(X(t))$ is decreasing in $t$. 
\end{lem}

\begin{cor}
Define the partial expectation operator $\pEx{X}{a}$ as follows: $\pEx{X}{a}:=\Ex f(X,a)$ where $f(x,a)=x1_{x\le a}+a1_{x > a}$. Then, if $X$ is IFR, $\pEx{X(t)}{a}$ is decreasing in $t$.
\end{cor}

\begin{remark}
dual of this problem...
\end{remark}

\begin{remark} \label{rem:origin}
The hazard rates of $A(t)$ and $C(t)$ at time $s$ (with respect to $t$) are equal to the hazard rates of $A(0)$, and $C(0)$ at time $t+s$, respectively. As a result, we have that Assumptions \ref{ass:relative}-\ref{ass:converging} also carry over to $A(t)$ and $C(t)$. This result is important in establishing results about the residual lifetime, as the time origin can be set arbitrarily.
\end{remark}

\begin{lem}
Let $e(t):=\Ex C_t$, and $l(t):=\Ex A_t$. Then Assumptions \ref{ass:converging}, and \ref{ass:IFR} imply that $l(t)-e(t)$ is decreasing in $t$.
\end{lem}
\begin{proof}
Consider a random variable $X(k,t)$ whose hazard rate function is as follows: 
\begin{align*}
\hrate{X(k,t)}{s}=
\begin{cases}
\hrate{A_t}{s} & s \le k\\
\hrate{C}{s} & o.w.
\end{cases}
\end{align*} 
Let $f(k):= \Ex X(k,t)$. Note that $X(0,t) =_{d} C_t$ (reza: justify). We show that $\diff{)}{arg2}$
\end{proof}
 [which means patients have better survival on either of these accesses at younger ages]
 \noat{We don't need identical distribution of $K$, or $M$. We can for instance assume a non-stationary distribution for them.}

\begin{prop}\label{prop:w'}
Assume $v(t,n,0)-v^{\pi_0}(t,n)$ is decreasing in $t$ in the interval $[t,t+a]$ for some $a>0$. Then if $v(t,n,0) \ge v^{\pi_0}(t,n)$, the optimal action in $[t,t+a]$ is to refer at $t$; Otherwise, no referral in $[t,t+a]$ can be optimal.
\end{prop}
\begin{proof}
Fix $t$, and $n$.  Assume that we take action $@_y$ for the current AVF chance, and then follow the optimal policy for the subsequent AVF chances. For $@_y$ to be an optimal action candidate, it is necessary that referral at $t+y$ is no worse than the null policy, i.e.
\begin{align}\label{eq:feasibility}
v(t+y,n,0) \ge v^{\pi_0}(t+y,n).
\end{align}
Now, we argue about the optimal action based on whether $v(t,n,0) \ge v^{\pi_0}(t,n)$ as follows:
\begin{itemize}
\item $v(t,n,0) \ge v^{\pi_0}(t,n)$: Note that referral at $t=0$ is NWN. For each candidate $@_y$, for which the Equation \ref{eq:feasibility} holds, we prove that $v(t,n,y)$ is decreasing in $y$ as follows.  Recall that based on Lemma \ref{lem:res'}, we have
\begin{align*} 
v(t,n,y)&=\surv{C_t}{y}\bigg[ v(t+y,n,0)-v^{\pi_0}(t+y,n)\bigg]+v^{\pi_0}(t,n)
\end{align*}
Thus $v(t,n,y)$ is decreasing in $y$, because it is the product of two non-negative decreasing functions of $y$, plus a constant. Thus, $y^*=0$, and referral at $t$ is the optimal action in $[t,t+a]$.
\item $v(t,n,0) < v^{\pi_0}(t,n)$: since $v(t,n,0) - v^{\pi_0}(t,n) \downarrow t$, there is no candidate $@_y$ that satisfy the necessary condition stated by Equation \ref{eq:feasibility}. Thus, no referral in $[t,t+a]$ can be optimal.
\end{itemize}
\end{proof}

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